dx-x-x-x-

Question Number 116701 by bemath last updated on 06/Oct/20

$$\int\:\frac{\mathrm{dx}}{\mathrm{x}+\mathrm{x}\sqrt{\mathrm{x}}}\:=? \\ $$

Commented by bobhans last updated on 06/Oct/20

∫ (dx/(x+x(√x) )) = ∫ (dx/(x(1+(√x) ))) [ letting x = λ^2 →dx = 2λ dλ ] I=∫ ((2λ dλ)/(λ^2 (1+λ))) = ∫ ((2 dλ)/(λ(1+λ))) = 2 [∫ ((1/λ) −(1/(1+λ)))dλ ] = 2 [ ln λ−ln (1+λ) ] + c = 2 ln (((√x)/(1+(√x))) ) + c

$$\:\int\:\frac{\mathrm{dx}}{\mathrm{x}+\mathrm{x}\sqrt{\mathrm{x}}\:}\:=\:\int\:\frac{\mathrm{dx}}{\mathrm{x}\left(\mathrm{1}+\sqrt{\mathrm{x}}\:\right)} \\ $$$$\left[\:\mathrm{letting}\:\mathrm{x}\:=\:\lambda^{\mathrm{2}} \:\rightarrow\mathrm{dx}\:=\:\mathrm{2}\lambda\:\mathrm{d}\lambda\:\right]\: \\ $$$$\mathrm{I}=\int\:\frac{\mathrm{2}\lambda\:\mathrm{d}\lambda}{\lambda^{\mathrm{2}} \left(\mathrm{1}+\lambda\right)}\:=\:\int\:\frac{\mathrm{2}\:\mathrm{d}\lambda}{\lambda\left(\mathrm{1}+\lambda\right)} \\ $$$$=\:\mathrm{2}\:\left[\int\:\left(\frac{\mathrm{1}}{\lambda}\:−\frac{\mathrm{1}}{\mathrm{1}+\lambda}\right)\mathrm{d}\lambda\:\right]\: \\ $$$$=\:\mathrm{2}\:\left[\:\mathrm{ln}\:\lambda−\mathrm{ln}\:\left(\mathrm{1}+\lambda\right)\:\right]\:+\:\mathrm{c} \\ $$$$=\:\mathrm{2}\:\mathrm{ln}\:\left(\frac{\sqrt{\mathrm{x}}}{\mathrm{1}+\sqrt{\mathrm{x}}}\:\right)\:+\:\mathrm{c} \\ $$

Commented by MJS_new last updated on 06/Oct/20

∫(dx/(x(1+x^q )))=−(1/q)ln ∣(1/x^q )+1∣ +C =ln ∣x∣ −(1/q)ln ∣x^q +1∣ +C

$$\int\frac{{dx}}{{x}\left(\mathrm{1}+{x}^{{q}} \right)}=−\frac{\mathrm{1}}{{q}}\mathrm{ln}\:\mid\frac{\mathrm{1}}{{x}^{{q}} }+\mathrm{1}\mid\:+{C}\:=\mathrm{ln}\:\mid{x}\mid\:−\frac{\mathrm{1}}{{q}}\mathrm{ln}\:\mid{x}^{{q}} +\mathrm{1}\mid\:+{C} \\ $$

Answered by Dwaipayan Shikari last updated on 06/Oct/20

∫(dx/( x+x(√x))) x=t^2 ⇒1=2t(dt/dx) ∫((2tdt)/(t^2 (1+t)))=2∫(1/(t(1+t)))dt=2log((t/(t+1)))+C =2log(((√x)/( (√x)+1)))+C

$$\int\frac{{dx}}{\:{x}+{x}\sqrt{{x}}}\:\:\:\:\:\:\:\:{x}={t}^{\mathrm{2}} \Rightarrow\mathrm{1}=\mathrm{2}{t}\frac{{dt}}{{dx}} \\ $$$$\int\frac{\mathrm{2}{tdt}}{{t}^{\mathrm{2}} \left(\mathrm{1}+{t}\right)}=\mathrm{2}\int\frac{\mathrm{1}}{{t}\left(\mathrm{1}+{t}\right)}{dt}=\mathrm{2}{log}\left(\frac{{t}}{{t}+\mathrm{1}}\right)+{C} \\ $$$$=\mathrm{2}{log}\left(\frac{\sqrt{{x}}}{\:\sqrt{{x}}+\mathrm{1}}\right)+{C} \\ $$

Commented by bemath last updated on 06/Oct/20

$$\mathrm{gaves}\:\mathrm{kudos} \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 06/Oct/20

I =∫ (dx/(x+x(√x))) ⇒ I =_((√x)=t) ∫ ((2tdt)/(t^2 +t^3 )) =2 ∫ (dt/(t+t^2 )) =2 ∫ (dt/(t(t+1))) =2 ∫((1/t)−(1/(t+1)))dt =2ln∣(t/(t+1))∣ +c =2ln∣((√x)/(1+(√x)))∣ +c

$$\mathrm{I}\:=\int\:\:\frac{\mathrm{dx}}{\mathrm{x}+\mathrm{x}\sqrt{\mathrm{x}}}\:\Rightarrow\:\mathrm{I}\:=_{\sqrt{\mathrm{x}}=\mathrm{t}} \:\:\:\int\:\:\frac{\mathrm{2tdt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{t}^{\mathrm{3}} }\:=\mathrm{2}\:\int\:\:\:\frac{\mathrm{dt}}{\mathrm{t}+\mathrm{t}^{\mathrm{2}} } \\ $$$$=\mathrm{2}\:\int\:\:\frac{\mathrm{dt}}{\mathrm{t}\left(\mathrm{t}+\mathrm{1}\right)}\:=\mathrm{2}\:\int\left(\frac{\mathrm{1}}{\mathrm{t}}−\frac{\mathrm{1}}{\mathrm{t}+\mathrm{1}}\right)\mathrm{dt}\:=\mathrm{2ln}\mid\frac{\mathrm{t}}{\mathrm{t}+\mathrm{1}}\mid\:+\mathrm{c} \\ $$$$=\mathrm{2ln}\mid\frac{\sqrt{\mathrm{x}}}{\mathrm{1}+\sqrt{\mathrm{x}}}\mid\:+\mathrm{c} \\ $$

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